Problem Description

In this problem, we are working with a graph that represents a system of cities and bidirectional roads connecting these cities. The graph is composed of n cities, which are numbered from 1 to n. The roads are represented by a 2D array roads, where each element in the array is a trio [a, b, distance]. This trio indicates that there is a road connecting city a and city b with a distance of distance. It's important to note that the graph of cities may not be fully connected, meaning some cities may not have a direct path to others.

The "score" of a path in this graph is a little unique; rather than summing up the total distance or considering average distance, the score is defined as the minimum distance of any road in that particular path. Put another way, out of all the roads used in a path, the score is the distance of the shortest one among them.

Our goal is to determine the minimum possible score of a path between city 1 and city n. A few rules apply to paths:

• A path is simply a sequence of roads connecting two cities; it may pass through intermediate cities.
• It is permissible for a path to include the same road more than once, if necessary.
• You can go through cities 1 and n multiple times to find the path with the optimal score.
• There is at least one path between city 1 and city n in all test cases.

Ultimately, we want to compute the highest "weakest" road in the best path from city 1 to city n.

Flowchart Walkthrough

To solve Leetcode 2492. Minimum Score of a Path Between Two Cities, let's use the Flowchart to identify the appropriate algorithm. Here's a detailed step-by-step analysis:

1. Is it a graph?

   • Yes: The problem involves cities and paths between them, which can naturally be represented as a graph where cities are nodes and paths are edges.

2. Is it a tree?

   • No: The graph described can have multiple paths between cities and is not necessarily a hierarchy, so it's not a tree.

3. Is the problem related to directed acyclic graphs (DAGs)?

   • No: The question does not specify that the graph is acyclic or directed. It's about finding a path in a general graph.

4. Is the problem related to shortest paths?

   • Yes: The objective is to find a path with the minimum score between two cities, which aligns with shortest path problems.

5. Is the graph weighted?

   • Yes: The problem involves scoring paths, implying that edges have weights.

6. Dijkstra's Algorithm

   • Since the graph is weighted and we're looking for the shortest path, Dijkstra's Algorithm becomes suitable for solving this problem. However, let's pretend this answer doesn't satisfy the condition to find a DFS-based approach. Ignore relying on Dijkstra's and continue on the flowchart for educational purposes.

7. Does the problem involve connectivity?

   • No: While the problem involves connections, the key problem is not just about connectivity but finding an optimized path, which shifts our focus away from generalized connectivity solutions.

The steps derived from applying the flowchart suggest the use of Dijkstra's Algorithm for a weighted shortest path problem. However, if explicitly asked to use Depth-First Search (or if using Dijkstra's is somehow unsuitable due to characteristics not detailed in the problem—say, negative weights or special conditions), one could modify DFS to handle priority decisions based on path scores. However, generally, Depth-First Search isn't inherently optimal for finding minimum path scores in weighted graphs without significant modifications. Hence, the flowchart directed us primarily towards Dijkstra's for this type of problem.

Intuition

To solve this problem, we can utilize Depth-First Search (DFS) to explore all possible paths from city 1 to city n. As we perform DFS, we track the minimum distance encountered along the current path. Before starting DFS, we initialize the answer ans with a value representing infinity (inf), because we will be minimizing this score as we go along.

To apply DFS, we first create a defaultdict of lists called g, which will be our graph where the keys are city numbers and the values are lists of tuples representing connected cities alongside the respective distances. We also create a list vis of boolean values, initialized to False, that keeps track of the cities that have been visited to prevent revisiting them in the current path exploration.

Our DFS process named dfs() takes a city i as an argument and iterates through all of its connections. For each adjacent city j with distance d to city i, we minimize ans with d to find the smallest distance seen so far along this path segment. We also mark j as visited and perform DFS from j. The process recursively continues until we have exhausted all possible paths.

After initializing our data structures, we simply call dfs(1) to start from the first city and allow the DFS to explore paths until city n. Once DFS is done, the ans variable will hold the minimum score among all paths from city 1 to city n.

The solution is neat because it elegantly uses DFS to explore the space of paths, and it cleverly updates the minimum score on-the-fly. It relies on the fact that, since the score is the minimum distance along a path, we just need to track the smallest road distance encountered at any point in our DFS exploration.

Learn more about Depth-First Search, Breadth-First Search, Union Find and Graph patterns.

Solution Approach

The implementation of the solution utilizes a Depth-First Search (DFS) algorithm which is a common strategy for exploring all the vertices of a graph. Let us walk through the important points of the implementation:

1. Graph Representation: A defaultdict of lists is used to represent the graph (g). Each key in the defaultdict represents a city, and the value is a list of tuples. Each tuple consists of a destination city and the distance to that city. This allows us to easily iterate over all neighbors of a city.

2. Visited Set: A list called vis is created with a size of n + 1, where n is the number of cities. The list is initialized with False to indicate that no city has been visited at the start. This ensures that we do not enter into an infinite loop by revisiting cities during our DFS traversal.

3. Distance Initialization: A variable ans is initialized with inf (infinity). This variable is used to keep track of the minimum distance encountered on any path between city 1 and city n.

4. DFS Algorithm: A recursive function dfs(i) is defined, which is responsible for traversing the graph:

   • Inside the DFS function, we iterate over each neighbor j and its associated distance d of the current city i.
   • We update ans with the minimum of its current value and d each time we encounter a smaller distance.
   • If the neighbor j has not yet been visited (vis[j] is False), we mark it as visited by setting vis[j] to True and then recursively call dfs(j) to continue exploring the graph from city j.

5. Starting the DFS Traversal: We call dfs(1) to begin our DFS traversal from city 1. This will explore different paths to reach city n.

6. Result: Given that there is at least one path from city 1 to city n, by the end of DFS traversal, ans will contain the minimum possible score (which is the minimum distance of the weakest road) in some path between city 1 and city n. This ans is then returned as a result.

The clever use of DFS allows for an efficient search through all the paths while keeping track of the minimum distance encountered, therefore the weakest link in terms of road distance, which effectively determines the score of the path.

Example Walkthrough

Let's illustrate the solution approach with a small example where n = 4 representing four cities and roads as [[1, 2, 4], [2, 3, 3], [3, 4, 2], [1, 3, 6]]. This setup implies the following connections:

• City 1 and city 2 are connected by a road with a distance of 4.
• City 2 and city 3 are connected by a road with a distance of 3.
• City 3 and city 4 are connected by a road with a distance of 2.
• City 1 and city 3 are directly connected as well with a distance of 6.

Following our solution approach:

1. Graph Representation: We first convert the roads array to a graph represented as a defaultdict of lists, g, so that we have:

   g = {
       1: [(2, 4), (3, 6)],
       2: [(1, 4), (3, 3)],
       3: [(2, 3), (4, 2), (1, 6)],
       4: [(3, 2)]
   }

   Each list contains tuples that represent connections to neighboring cities and the distance to them.

2. Visited Set: We create vis = [False, False, False, False, False].

3. Distance Initialization: Set ans = inf.

4. DFS Algorithm: For the dfs function, starting with dfs(1):

   • Look at neighbors of city 1: (2, 4) and (3, 6).
   • For neighbor (2, 4), since city 2 is not visited, mark it visited and compare ans with 4. ans becomes min(inf, 4) = 4.
   • Continue DFS by calling dfs(2).
     • In dfs(2), iterate over neighbors: (1, 4) and (3, 3).
     • Skip city 1 since it's visited; for city 3 with distance 3, since city 3 is not visited, mark it visited and update ans to min(4, 3) = 3.
     • Continue DFS by calling dfs(3).
       • In dfs(3), check neighbors (2, 3), (4, 2), and (1, 6).
       • City 2 is visited, skip. Update ans with min(3, 2) when visiting city 4, so ans becomes 2.
       • City 4 has no unvisited neighbors left, so DFS goes back up.

   This process continues until all possible paths between city 1 and city 4 have been explored through recursion.

5. Result: After the DFS is finished, ans holds the minimum distance encountered on any path, which in this case is 2. That means the highest "weakest" road on the best path from city 1 to city 4 has a distance of 2. This is the score of our path.

Hence, the minimum possible score for a path between city 1 and city 4 is 2, given this set of roads and cities.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code implements a depth-first search (DFS) on a graph represented as a adjacency list. The dfs function is called recursively to traverse the graph.

The time complexity of the algorithm depends on the number of nodes n and the number of edges in the roads list. The DFS will visit each node exactly once, due to the vis[j] = True guard before each recursive call.

The complexity can have two major components: the time it takes to set up the adjacency list and the time it takes to do the DFS traversal.

• Adjacency List Creation: The adjacency list is created by iterating through each road in the roads list. This operation takes O(E) time where E is the number of edges as each edge is visited once.

• DFS Traversal: DFS traversal typically has a time complexity of O(V + E) for visiting each node and edge at most once.

Therefore the total time complexity is O(V + E) where V is the number of vertices and E is the number of edges.

Space Complexity

The space complexity includes the storage for the graph (adjacency list), the visited array, and the stack space used by the recursive calls of DFS.

• Graph Storage: Each node stores a list of its edges, so in total this is proportional to O(E), where E is the number of edges.

• Visited Array: This is an array of length n + 1, thus taking up O(V) space where V is the number of vertices.

• DFS Stack space: In the worst case, the recursive DFS could go as deep as the number of nodes in the graph (imagine a linked-list shaped graph), so the stack space in the worst case can be O(V).

Consequently, the total space complexity is O(E + V) with consideration for the adjacency list, visited array, and recursive stack space.

Learn more about how to find time and space complexity quickly using problem constraints.